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How many 19th power residues modulo 229 are there?

My approach was that as gcd(19,228) = 19, then by a proposition that I am allowed to use from lecture, any b is a 19th power residue mod 229 if and only if $b = x^{19}$ for some x an element of unit group of (Z/229). This group has 228 elements, so I conclude that there are 228 19th power residues modulo 229.

Is my logic flawed in any sense? I am not sure if the fact I am using necessarily implies a bijection between the number of power residues and elements in the unit group.

Bill Dubuque
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Anon
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  • Careful...$19,|,228$ so there are non-trivial solutions to $a^{19}\equiv 1 \pmod {229}$. – lulu Oct 06 '23 at 17:38
  • Interesting, how would one go about finding the non-trivial 19th power residues? – Anon Oct 06 '23 at 17:42
  • You could just search, but there's no need. The goal is to count them, not list them. – lulu Oct 06 '23 at 17:42
  • I suggest: start with a smaller example. How many cubes are there $\pmod {13}$? Try to generalize that. – lulu Oct 06 '23 at 17:44
  • Here is the proposition shown in lecture: If (n,p-1) = e, then some b is nth power residue mod p if and only if b = x^e for some x in unit group of Z/p. Doesn't this imply that the solutions MUST be elements of the unit group? – Anon Oct 06 '23 at 17:47
  • Of course the solutions are units, by definition. You can add $0$ as a power if you want, but I assumed you were talking about non-zero powers. Work my example $\pmod {13}$. that's small enough that you can easily do it by hand and it illustrates the general pattern. – lulu Oct 06 '23 at 17:48
  • Does this answer your question? Cyclic Group Generators of Order $n$ – Mike Oct 06 '23 at 17:57
  • A brute-force computer solution gives the answer as 13 — 0, 1, 18, 89, 94, 95, 107, 122, 134, 135, 140, 211, and 228. – Dan Oct 06 '23 at 18:39

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$229$ is prime so $(\mathbb Z/229)^×$ is cyclic group and the collection of all $19$th powers forms subgroup, just find its order. I don't know if such simple answer exists if it wasn't a prime.

Tony Pizza
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  • The subgroup is exactly the set of all elements of order $12=228/19$ and so has order $12$. See https://math.stackexchange.com/a/410464/589 – lhf Oct 08 '23 at 10:35